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Hooke's law and angular freq.

  1. Aug 19, 2005 #1
    hmm ok, i was watching the MIT opencourseware video on oscillations and there was a part where it was mentioned that,

    the diff. eq. [tex] x''+ \frac {k}{m} x = 0 [/tex] has solution [tex] x= x_0 cos (\omega t + \phi) [/tex] if and only if [tex]\omega= \sqrt{\frac {k}{m}} [/tex]

    how do i show that omega is the sqaureoot of k over m? thanks alot.
  2. jcsd
  3. Aug 19, 2005 #2


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    Differentiate the solution twice with respect to time and substitute it back into the differential equation - then tell us what you discovered! :)
  4. Aug 19, 2005 #3
    ohhh, i see, it can be gotten by verifying the solution. I thought we needed to do something to hooke's law, but verifying the solution works great too.

    : )

    edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that [tex]\omega= \sqrt{\frac {k}{m}} [/tex]. How do we know that [tex]\omega= \sqrt{\frac {k}{m}} [/tex] when we are solving it?
    Last edited: Aug 19, 2005
  5. Aug 19, 2005 #4


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    If you just substitute [itex]x = x_0 \cos(\omega t - \phi)[/itex] into the differential equation the required value of [itex]\omega[/itex] will jump out at you!
  6. Aug 19, 2005 #5


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    In this case, [tex]\omega[/tex] is referring to the angular frequency, which is the multiplicative factor in front of the independent variable in the sine or cosine function. It's just a standard substitution that they probably just didn't bother to define.
  7. Aug 21, 2005 #6
    try the following:

    in your original ODE, consider [itex] \omega ^2[\itex] just as a mathematically sound way to express the positiveness of the factor [itex] \frac{k}{m} [\itex]. After all the calculations you will realize that this choice has proven itself useful and physically meaningful.
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